Question

An ellipse has vertices along the major axis at (0, 1) and (0, −9). The foci of the ellipse are located at (0, −1) and (0, −7). The equation of the ellipse is in the form below.

Answer 1

so, with those points provided, notice the vertices are lying along the y-axis, check the picture below, thus is a vertical ellipse.

now, the center is half-way between the vertices, therefore it'd be at 0, -4, like in the picture in red.

the distance from the center to either foci, is "c", and that's c = 3.

the "a" component of the major axis is 5 units, now let's find the "b" component,

$\bf \textit{ellipse, vertical major axis} \\\\ \cfrac{(x- h)^2}{ b^2}+\cfrac{(y- k)^2}{ a^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h, k\pm a)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2- b ^2} \end{cases}\\\\ -------------------------------$

$\bf \begin{cases} a=5\\ c=3 \end{cases}\implies c=\sqrt{a^2-b^2}\implies c^2=a^2-b^2\implies b^2=a^2-c^2 \\\\\\ b=\sqrt{a^2-c^2}\implies b=\sqrt{5^2-3^2}\implies b=4\\\\ -------------------------------\\\\ \begin{cases} h=0\\ k=-4\\ a=5\\ b=4 \end{cases}\implies \cfrac{(x- 0)^2}{ 4^2}+\cfrac{[y-(-4)]^2}{ 5^2}=1 \\\\\\ \cfrac{(x- 0)^2}{ 16}+\cfrac{(y+4)^2}{25}=1$

Answer 2

The equation of the ellipse is required.

The required equation is $\dfrac{x^2}{16}+\dfrac{(y+4)^2}{25}=1$

It can be see that the major axis is parallel to the y axis.

The major axis points are

$(h,k+a)=(0,1)$

$(h,k-a)=(0,-9)$

$k+a=1$

$k-a=-9$

Subtracting the equations

$2a=10\\\Rightarrow a=5$

The foci are

$(h,k+c)=(0,-1)$

$(h,k-c)=(0,-7)$

$k+c=-1$

$k-c=-7$

Subtracting the equations

$2c=6\\\Rightarrow c=3$

$k+c=-1\\\Rightarrow k=-1-c\\\Rightarrow k=-1-3\\\Rightarrow k=-4$

Foci is given by

$c^2=a^2-b^2\\\Rightarrow b=\sqrt{a^2-c^2}\\\Rightarrow b=\sqrt{5^2-3^2}=4$

The equation is

$\dfrac{(x-h)^2}{b^2}+\dfrac{(x-k)^2}{a^2}=1\\\Rightarrow \dfrac{(x-0)^2}{4^2}+\dfrac{(y+4)^2}{5^2}=1\\\Rightarrow \dfrac{x^2}{16}+\dfrac{(y+4)^2}{25}=1$

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